Question
Show that the point $(11, – 2)$ is equidistant from $(4, – 3)$ and $(6, 3)$
✓
Answer
Let $P\left(x_1, y_1\right)=P(11,-2), Q\left(x_2, y_2\right)=Q(4,-3), R\left(x_3, y_3\right)=R(6,3)$
By distance formula,
$d(P, Q)=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
$=\sqrt{(4-11)^2+[-3-(-2)]^2}$
$=\sqrt{(-7)^2+(-1)^2}$
$=\sqrt{49+1}$
$=\sqrt{50}$
$=5 \sqrt{2} $
And
$ d(P, R)=\sqrt{\left(x_3-x_1\right)^2+\left(y_3-y_1\right)^2}$
$=\sqrt{(6-11)^2+[3-(-2)]^2}$
$=\sqrt{(-5)^2+(5)^2}$
$=\sqrt{25+25}$
$=\sqrt{50}$
$=5 \sqrt{2}$
Here, $d(P, Q)=d(P, R)$
$\therefore$ Point $(11,-2)$ is equidistant from $(4,-3)$ and $(6,3)$.
By distance formula,
$d(P, Q)=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
$=\sqrt{(4-11)^2+[-3-(-2)]^2}$
$=\sqrt{(-7)^2+(-1)^2}$
$=\sqrt{49+1}$
$=\sqrt{50}$
$=5 \sqrt{2} $
And
$ d(P, R)=\sqrt{\left(x_3-x_1\right)^2+\left(y_3-y_1\right)^2}$
$=\sqrt{(6-11)^2+[3-(-2)]^2}$
$=\sqrt{(-5)^2+(5)^2}$
$=\sqrt{25+25}$
$=\sqrt{50}$
$=5 \sqrt{2}$
Here, $d(P, Q)=d(P, R)$
$\therefore$ Point $(11,-2)$ is equidistant from $(4,-3)$ and $(6,3)$.
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